Choose the correct answer from the given four options:
The $ 4^{\text {th }} $ term from the end of the AP: $ -11,-8,-5, \ldots, 49 $ is
(A) 37
(B) 40
(C) 43
(D) 58
Given:
Given A.P. is \( -11,-8,-5, \ldots, 49 \).
To do:
We have to find the 4th term from the end of the given arithmetic progression.
Solution:
In the given A.P.,
$a_1=-11, a_2=-8, a_3=-5$
First term $a_1 = a= -11$, last term $l = 49$
Common difference $d = a_2-a_1 = -8 - (-11) = -8+11=3$
We know that,
nth term from the end is given by $l - (n - 1 ) d$.
Therefore,
4th term from the end $= 49 - (4 - 1) \times (3)$
$= 49 - 3 \times 3$
$= 49 -9$
$= 40$.
The 4th term from the end of the given A.P. is $40$.
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